Final answer:
The foci of the given ellipse are located at (4, 2) and (4, 6).
Step-by-step explanation:
An ellipse can be expressed in the form \(\frac{{(x-h)^2}}{{a^2}} + \frac{{(y-k)^2}}{{b^2}} = 1\), where \((h,k)\) represents the center of the ellipse, \(a\) represents the semi-major axis, and \(b\) represents the semi-minor axis.
To find the foci of the given ellipse, we need to first rearrange the equation in standard form. We can do this by grouping the \(x\) terms and the \(y\) terms:
\[9x^2 - 72x + 5y^2 + 99 = 0\]
To complete the square for both \(x\) and \(y\), we divide the \(x\) terms by 9 and the \(y\) terms by 5:
\[\frac{{x^2 - 8x}}{9} + \frac{{y^2}}{5} + \frac{99}{5} = 0\]
Next, we add and subtract the appropriate constant values to complete the square for \(x\) and \(y\):
\[\frac{{x^2 - 8x + 16}}{9} - \frac{{16}}{9} + \frac{{y^2}}{5} - \frac{{16}}{5} + \frac{99}{5} = 0\]
Combining like terms:
\[\frac{{(x-4)^2}}{9} + \frac{{(y-\sqrt{16})^2}}{5} + \frac{{99-80}}{5} = 0\]
Simplifying further:
\[\frac{{(x-4)^2}}{9} + \frac{{(y-\sqrt{16})^2}}{5} + \frac{19}{5} = 0\]
Now, we can see that the center of the ellipse is at \((4, \sqrt{16}) = (4, 4)\) and the semi-major axis is \(a = 3\) and the semi-minor axis is \(b = \sqrt{5}\).
The distance from the center to each focus is given by the formula \(c = \sqrt{a^2 - b^2}\).
Substituting the values of \(a\) and \(b\) into the formula:
\[c = \sqrt{3^2 - \sqrt{5}^2} = \sqrt{9 - 5} = \sqrt{4} = 2\]
So, the foci of the given ellipse are 2 units away from the center along the major axis. The foci are located at \((4, 4-2) = (4, 2)\) and \((4, 4+2) = (4, 6)\).